The relation ${R_1}$ is defined for $a,b \in {\mathbb{Z}^ + }$ by $a{R_1}b$ if and only if $n\left| {({a^2} - {b^2})} \right.$ where $n$ is a fixed positive integer.

  (i)     Show that ${R_1}$ is an equivalence relation.

  (ii)     Determine the equivalence classes when $n = 8$ .

Consider the group $\left\{ {G, * } \right\}$ and let $H$ be a subset of $G$ defined by

$H = \left\{ {x \in G} \right.$ such that $x * a = a * x$ for all $a \in \left. G \right\}$ .

Show that $\left\{ {H, * } \right\}$ is a subgroup of $\left\{ {G, * } \right\}$ .

The relation ${R_2}$ is defined for $a,b \in {\mathbb{Z}^ + }$ by $a{R_2}b$ if and only if $(4 + \left| {a - b} \right|)$ is the square of a positive integer. Show that ${R_2}$ is not transitive.

(i)     Since ${a^2} - {a^2} = 0$ is divisible by n, it follows that $a{R_1}a$ so ${R_1}$ is reflexive.     A1

$a{R_1}b \Rightarrow {a^2} - {b^2}$ divisible by $n \Rightarrow {b^2} - {a^2}$ divisible by $n \Rightarrow b{R_1}a$ so

symmetric.     A1

$a{R_1}b$ and $b{R_1}c \Rightarrow {a^2} - {b^2} = pn$ and ${b^2} - {c^2} = qn$    A1

$({a^2} - {b^2}) + ({b^2} - {c^2}) = pn + qn$     M1

so ${a^2} - {c^2} = (p + q)n \Rightarrow a{R_1}c$     A1

Therefore ${R_1}$ is transitive.

It follows that ${R_1}$ is an equivalence relation.     AG

(ii)     When $n = 8$ , the equivalence classes are

$\left\{ {1,3,5,7,9, \ldots } \right\}$ , i.e. the odd integers     A2

$\left\{ {2,6,10,14, \ldots } \right\}$     A2

and $\left\{ {4,8,12,16, \ldots } \right\}$     A2

Note: If finite sets are shown award A1A1A1.

[11 marks]

Associativity follows since G is associative.     A1

Closure: Let $x,y \in H$ so $ax = xa$ , $ay = ya$ for $a \in G$     M1

Consider $axy = xay = xya \Rightarrow xy \in H$     M1A1

The identity $e \in H$ since $ae = ea$ for $a \in G$     A2

Inverse: Let $x \in H$ so $ax = xa$ for $a \in G$

Then

${x^{ - 1}}a = {x^{ - 1}}ax{x^{ - 1}}$     M1A1

$= {x^{ - 1}}xa{x^{ - 1}}$     M1

$= a{x^{ - 1}}$     A1

so $\Rightarrow {x^{ - 1}} \in H$     A1

The four group axioms are satisfied so $H$ is a subgroup.     R1

[12 marks]

Attempt to find a counter example.     (M1)

We note that $1{R_2}6$ and $6{R_2}11$ but 1 not ${R_2}11$ .     A2

Note: Accept any valid counter example.

The relation is not transitive.     AG

[3 marks]